The proof uses complex interpolation theory essentially in the same way as in the proof of the classical Hausdorff-Young inequality (see, e.g., [1]}). For background on interpolation theory we refer to the short survey of Connes [2]} and the references therein. The main reference for complex interpolation theory there is the article of Calderón [3]} (see also [1]}, [5]}).
As mentioned above \(L_2(n_\theta )\)  and \(\hat{\ell }_2(\theta ^n)\)  agree as Banach spaces and the inclusion of \(\ell _1(n_\theta )\)  into \(C(n_\theta )\)  gives rise to a contraction into \(L_\infty (n_\theta )\) . In particular, we have the inequalities,
\(\Vert x\Vert _{L_2}=\Vert x\Vert _{\hat{\ell }_2} \qquad \forall x\in \hat{\ell }_2(n_\theta ) ,\\\Vert x\Vert _{L_\infty }\le \Vert x\Vert _{\hat{\ell }_1} \qquad \forall x\in \hat{\ell }_1(n_\theta ). \) 
Suppose that \(q\in (2,\infty )\) . Then \(L_{q}(\theta ^n)\)  is a complex interpolation space for the pair of Banach spaces \((L_2(n_\theta ), L_\infty (n_\theta ))\) . Namely, in the notation of [2]} we have \(L_q(\theta ^n)= [L_2(n_\theta ), L_\infty (n_\theta )]_\theta \) , where \(\theta \)  is such that \(q=(1-\theta )\frac{1}{2}\) , i.e., \(\theta =1-2q^{-1}\)  (see [7]} and [8]}). Note also that, as \(r\in (1,2)\) , the space \(\ell _r(n_\theta )\)  is an exact interpolation space for the pair \((\ell _2(n_\theta ),\ell _1(n_\theta ))\) . Namely, \(\ell _r(n_\theta )=[\ell _2(n_\theta ),\ell _1(n_\theta )]_\theta \) , since we have
\((1-\theta )\frac{1}{2}+\theta =\frac{1}{2}(1+\theta )=\frac{1}{2}(2-2q^{-1})=1-q^{-1}=r^{-1}.\) 
As the Fourier transform induces isometric isomorphisms between the spaces \(\ell _p(\mathbb {Z}^n)\)  and \(\hat{\ell }_p(n_\theta )\) , we see that
\(\hat{\ell }_r(n_\theta )=[\hat{\ell }_2(n_\theta ),\hat{\ell }_1(n_\theta )]_\theta \) . Combining all this with the inequalities (REF )–() and using complex interpolation theory (see [2]}) then shows we have a continuous inclusion \(\hat{\ell }_r(n_\theta )\subseteq L_{q}(n_\theta )\)  with the norm inequality (REF ). The proof is complete.
